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<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_3_1.html ./knowl/eq3_4_1.html">
\begin{equation*}
\begin{aligned}
&amp;y(0)=2:\quad 2=C_1+C_2,\\
&amp; y^{\prime}(0)=-1:\quad -1=C_1-C_2,
&amp;~\rightarrow~C_1=\frac{1}{2},\quad C_2=\frac{3}{2}.
\end{aligned}
\end{equation*}
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<p class="continuation">Thus,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq3_3_1.html ./knowl/eq3_4_1.html">
\begin{equation*}
y=\frac{1}{2} e^x+\frac{3}{2} e^{-x}.
\end{equation*}
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<p class="continuation">This is the solution to the initial value problem of (<a href="" class="xref" data-knowl="./knowl/eq3_3_1.html" title="Equation 3.1.4">(3.1.4)</a>) and (<a href="" class="xref" data-knowl="./knowl/eq3_4_1.html" title="Equation 3.1.5">(3.1.5)</a>).</p>
<span class="incontext"><a href="sec3_1.html#p-62" class="internal">in-context</a></span>
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